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Question Number 119517 by bemath last updated on 25/Oct/20
limx→∞[sin(x+1x)−sinx]=?
Answered by Dwaipayan Shikari last updated on 25/Oct/20
limx→∞(sin(x+1x)−sinx)=limx→∞(2cos(x+12x)sin(1x))=0−1⩽cos(x+12x)⩽1limx→∞sin1x→0
Answered by benjo_mathlover last updated on 25/Oct/20
limx→∞[sinxcos1x+cosxsin1x−sinx]=limx→∞[sinx(cos1x−1)+cosxsin1x][notelimx→∞cos1x−1=0∧limx→∞cosxsin1x=0becausecosxoscilatesbetween−1and1]thuslimx→∞[sinx(cos1x−1)+cosxsin1x]=0
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